Find a parabola which equation is $y=ax^2+bx$ and its tangent in $(1,1)$ is the straight line $y=3x-2$ I would be glad to have some hints or clues in this problem, I have tried doing the derivate of $y$ which is $2ax+b$ but I do not know how does this help

Simpler: write the abscissæ equation for the intersection of the parabola and the line: $$ax^2+bx=3x-2\iff ax^2+(b-3)x+2=0.$$ The line is tangent to the parabola if and only if this quadratic equation has a double root, and the point of contact is $(1,1)$ if and only if the double root is $x=1$.

Can you take it from here?